Question

A.) For which reaction will K_p = K_c?

	2 H2O(l) ↔ 2 H2(g) + O2(g)
	2 HgO(s) ↔ Hg(l) + O2(g)
	S(s) + O2(g) ↔ SO2(g)
	H2CO3(s) ↔ H2O(l) + CO2(g)
	CaCO3(s) ↔ CaO(s) + CO2(g)

B.) How does an increase in the pressure within the container by the addition of the inert gas E affect the amount of C present at equilibrium for the following reaction?
A(g) + 3 B(g) ↔ 2 C(g) + D(g)           ΔH° = -65 kJ

	The amount of C will increase.
	The amount of C will not change.
	Cannot be predicted.
	The amount of C will decrease.

C.) Which statement is true for the following reaction at equilibrium at 25°C?

H₂(g) + Br₂(g) ⇌ 2HBr(g); K = 1.9 × 10¹⁹

	Concentration of HBr is low.
	The forward reaction does not happen.
	The rate of the reaction is very slow.
	Concentration of H2 and Br2 are low.

D.) If the temperature is increased and some oxygen is removed, how does this affect the amount of the Cl₂ present at equilibrium for the following reaction?

4 HCl(g) + O₂(g) ↔ 2 H₂O(g) + 2 Cl₂(g)           ΔH° = 120 kJ

	Cannot be predicted.
	The amount of Cl2 will not change.
	The amount of Cl2 will increase.
	The amount of Cl2 will decrease.

Answer 1

A.

K_p will equal K_c only when the moles of gas is the same on both sides of the equation. The only one for which that is the case is

S(s) + O₂(g) ↔ SO₂(g)

B.

The amount of C will not change.

Adding an inert gas into a gas-phase equilibrium at constant volume does not result in a shift. This is because the addition of a non-reactive gas does not change the partial pressures of the other gases in the container. While the total pressure of the system increases, the total pressure does not have any effect on the equilibrium constant.

C.

Concentration of H₂ and Br₂ are low.

Since,

K= 1.9 × 10¹⁹ = [HBr]² / ([H₂][Br₂])

D.

Cannot be predicted.

4 HCl(g) + O₂(g) ↔ 2 H₂O(g) + 2 Cl₂(g)           ΔH° = 120 kJ

Temperature increased - Endothermic reaction - forward - Cl₂ concentration will increase.

oxygen is removed - Equilibirium will shift to backward direction.